Mathematics (Basic) Test 2

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Mathematics (Basic) Test 2

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Question 1
1 / -0

Given that sin \(\theta\) =a/b ,then tan \(\theta\) is equal to

A
\(\frac{b}{\sqrt{a^2+b^2}}\)

B
\(\frac{b}{\sqrt{b^2-a^2}}\)

C
\(\frac{a}{\sqrt{a^2-b^2}}\)

D
\(\frac{a}{\sqrt{b^2-a^2}}\)

Solution

sin \(\theta\) =\(\frac{a}{b}\)

\(H^2=P^2+B^2\)

\(b^2=a^2+B^2\)

B=\(\sqrt{(b^2-a^2)}\)

tan \(\theta\) =\(\frac{P}B\) = \(\frac{a}{\sqrt{(b^2-a^2)}}\)
Question 2
1 / -0

If x = \(2sin^2\theta\) and y = \(2cos^2\theta\) + 1 then x + y is

A
3

B
2

C
1

D
\(\frac12\)

Solution

x + y = \(2sin^2\theta+2cos^2\theta+1\)

= \(2(sin^2\theta + cos^2\theta )+1\)

= 2 + 1 = 3
Question 3
1 / -0

If the difference between the circumference and the radius of a circle is 37cm ,\(\pi\) = 22/7, the circumference (in cm) of the circle is

A
154

B
44

C
14

D
7

Solution

2\(\pi\)r- r = 37 r{2\(\times\)(\(\frac{22}7\))-1} = 37

r= 37\(\times\frac{7}{37}\)

r = 7

circumference=2\(\times\)(\(\frac{22}7\)) \(\times\) 7 = 44 cm
Question 4
1 / -0

The least number that is divisible by all the numbers from 1 to 10 (both inclusive)

A
100

B
1000

C
2520

D
5040

Solution

1 = 1

2 = 2 \(\times\) 1

3 = 3 \(\times\) 1

4 = 2 \(\times\) 2

5 = 5 \(\times\) 1

6 = 2 \(\times\) 3

7 = 7 \(\times\) 1

8 = 2 \(\times\) 2 \(\times\) 2

9 = 3 \(\times\) 3

10 = 2 \(\times\) 5

So, LCM of these numbers = 1 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 3 \(\times\) 5 \(\times\) 7 = 2520

Hence, least number divisible by all the numbers from 1 to 10 is 2520
Question 5
1 / -0

Three bells ring at intervals of 4, 7 and 14 minutes. All three rang at 6 AM. When will they ring together again?

A
6:07 AM

B
6:14 AM

C
6:28 AM

D
6:25 AM

Solution

LCM of 4,7,14 = 28

Bells will they ring together again at 6:28 AM
Question 6
1 / -0

What is the age of father, if the sum of the ages of a father and his son in years is 65 and twice the difference of their ages in years is 50?

A
40 years

B
45 years

C
55 years

D
65 years

Solution

Let age of Father = x Years

Let age of son = y years

x + y = 65

2(x - y) = 50

Solving the above equations

Father’s Age =x = 45 years
Question 7
1 / -0

What is the value of \((tan\theta\,cosec\theta)^2 -(sin\theta\,sec\theta)^2\)

A
-1

B
0

C
1

D
2

Solution

\((tan\theta\,cosec\theta)^2 -(sin\theta\,sec\theta)^2\)

\(=tan^2\theta \,cosec^2\theta-sin^2\theta\,sec^2\theta\)

\(=(\frac{sin^2\theta}{cos^2\theta})\times\frac1{ sin^2\theta} \)\(- sin^2\theta \times\frac1{cos^2\theta}\)

\(=\frac{(1- sin^2\theta)}{ cos^2\theta}= \frac{cos^2\theta}{ cos^2\theta} =1\)
Question 8
1 / -0

The perimeters of two similar triangles are 26 cm and 39 cm.The ratio of their areas will be

A
2 : 3

B
6 : 9

C
4 : 6

D
4 : 9

Solution

\(\frac{A1}{A2}=(\frac{P1}{P2})^2=(\frac{26}{39})^2\)

\(\frac{A1}{A_2}=(\frac23)^2=\frac49\)
Question 9
1 / -0

There are 20 vehicles-cars and motorcycles in a parking area. If there are 56 wheels together, how many cars are there?

A
8

B
10

C
12

D
20

Solution

Let no of Cars=x

Let no of motorcycles = y

X + y = 20

4x + 2y = 56

Solving the above equations

No of cars = x = 8
Question 10
1 / -0

A man goes 15m due west and then 8m due north. How far is he from the starting point?

A
7m

B
10m

C
17m

D
23m

Solution

\(H^2=P^2+B^2\)

\(H^2=15^2+8^2\)

H=17m
Question 11
1 / -0

What is the length of an altitude of an equilateral triangle of side 8cm?

A
2\(\sqrt3\) cm

B
3\(\sqrt3\) cm

C
4\(\sqrt3\) cm

D
5\(\sqrt3\) cm

Solution

\((altitude)^2=(side)^2 -(\frac{side}2)^2\)

=\(8^2 -4^2\) = 64 - 16 = 48

Altitude=4 \(\sqrt3\) cm
Question 12
1 / -0

If the letters of the word RAMANUJAN are put in a box and one letter is drawn at random. The probability that the letter is A is

A
\(\frac35\)

B
\(\frac12\)

C
\(\frac37\)

D
\(\frac13\)

Solution

P = \(\frac39=\frac13\)
Question 13
1 / -0

Area of a sector of a circle is \(\frac16\) to the area of circle. Find the degree measure of its minor arc

A
\(90^{\circ}\)

B
\(60^{\circ}\)

C
\(45^{\circ}\)

D
\(30^{\circ}\)

Solution

\(\frac{\theta}{360^{\circ}}\times \pi r^2=\frac1{6}\times \pi r^2\)

\(\theta=60^{\circ}\)
Question 14
1 / -0

A vertical stick 20m long casts a shadow 10m long on the ground. At the same time a tower casts a shadow 50m long. What is the height of the tower?

A
30m

B
50m

C
80m

D
100m

Solution

\(\frac{Height\,of\,Vertical\,stick}{Shadow\,of\,vertical\,stick}\)\(=\frac{height\,of\,tower}{shadow\,of\,tower}\)

\(\frac{20}{10}=\frac{Height\,of\,tower}{50}\)

Height of tower = 100 m
Question 15
1 / -0

What is the solution of the pair of linear equations 37x + 43y=123, 43x + 37y=117?

A
x = 2, y = 1

B
x = -1, y = 2

C
x = -2, y = 1

D
x = 1, y = 2

Solution

37x + 43y = 123 ____(1)

43x + 37y = 117 ____(2)

Adding (1) and (2)

X + y = 3 ______(3)

Subtracting (2) from (1)

-x + y = 1..............(4)

Adding (3) and (4),

2y = 4

y = 2

⇒ x = 1

\(\therefore\) solution is x = 1 and y = 2
Question 16
1 / -0

The distance between the point Country A and Country B is

A
4 units

B
5 units

C
6 units

D
7 units

Solution

AB=\(\sqrt{(4-1)^2+(0-4)^2}\)

= \(\sqrt{(3^2+4^2 )}\)

AB = 5 units
Question 17
1 / -0

Find a relation between x and y such that the point (x,y) is equidistant from the Country C and Country D

A
x - y = 2

B
x + y = 2

C
2x - y = 0

D
2x + y = 2

Solution

\((\text x-7)^2+(y-1)^2=(\text x-3)^2+(y-5)^2\)

\(X^2+49-14\text x+y^2+1-2y\)

\(=\text x^2+9-6\text x+y^2+25-10y\)

Simplifying

x - y = 2
Question 18
1 / -0

The fault line 3x + y – 9 = 0 divides the line joining the Country P(1, 3) and Country Q(2, 7) internally in the ratio

A
3 : 4

B
3 : 2

C
2 : 3

D
4 : 3

Solution

3x + y – 9 = 0

Let R divide the line in ratio k : 1

\(R\left(\frac{ 2k+1}{k+1}, \frac{7k+3}{k+1}\right)\)

\(3\left(\frac{2k+1}{k+1}\right)+\left(\frac{7k+3}{k+1}\right)-9=0\)

4k - 3 = 0

K = \(\frac34\)

3 : 4
Question 19
1 / -0

The distance of the Country M from the x-axis is

A
1 units

B
2 units

C
3 units

D
5 units

Solution

Distance of M from X - axis

= \(\sqrt{(2-2)^2+(0-3)^2}\)

= \(\sqrt9\) = 3 units
Question 20
1 / -0

What are the co-ordinates of the Country lying on the mid-point of Country A and Country D?

A
(1, 3)

B
(2, \(\frac92\))

C
(4, \(\frac5 2\))

D
(\(\frac9 2\), 2)

Solution

\(\left(\frac{(1+3)}{2} , \frac{(4+5)}2\right)\)

\(= (\frac42, \frac92) = (2, \frac92)\)
Question 21
1 / -0

If the Roller Coaster is represented by the cubic polynomial \(t(\text x)=p\text x^3+q\text x^2+r\text x+s\), then which of the following is always true

A
\(s\neq0\)

B
\(r\neq0\)

C
\(q\neq0\)

D
\(p\neq0\)

Solution

\(p\neq0\)

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Mathematics (Basic) Test 2

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Mathematics (Basic) Test 2

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SHARING IS CARING

Question 1

\(\frac{b}{\sqrt{a^2+b^2}}\)

\(\frac{b}{\sqrt{b^2-a^2}}\)

\(\frac{a}{\sqrt{a^2-b^2}}\)

\(\frac{a}{\sqrt{b^2-a^2}}\)

Solution

Question 2

3

2

1

\(\frac12\)

Solution

Question 3

154

44

14

7

Solution

Question 4

100

1000

2520

5040

Solution

Question 5

6:07 AM

6:14 AM

6:28 AM

6:25 AM

Solution

Question 6

40 years

45 years

55 years

65 years

Solution

Question 7

-1

0

1

2

Solution

Question 8

2 : 3

6 : 9

4 : 6

4 : 9

Solution

Question 9

8

10

12

20

Solution

Question 10

7m

10m

17m

23m

Solution

Question 11

2\(\sqrt3\) cm

3\(\sqrt3\) cm

4\(\sqrt3\) cm

5\(\sqrt3\) cm

Solution

Question 12

\(\frac35\)

\(\frac12\)

\(\frac37\)

\(\frac13\)

Solution